A problem of Classical Mechanics by Swapnil Das

Classical Mechanics Level 4

A ball of mass m m is dropped from a height h 1 h_1 , and after striking the ground the ball bounces back to a height h 2 h_2 ( h 1 > h 2 h_1>h_2 ).

At a height h = 1 2 h 2 h=\frac12 h_2 , the ratio of the kinetic energy during the initial drop to that during the bounce is 2:1, what is the ratio of h 1 : h 2 h_1:h_2 ?


The answer is 1.5.

Swapnil Das by Swapnil Das , 20, India

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1 solution

Brilliant Physics Staff
Oct 1, 2015

For the first ball, in dropping from the height h 1 h_1 to 1 2 h 2 \frac12 h_2 , the ball picks up the speed 2 g Δ h \sqrt{2g\Delta h} , thus its kinetic energy at the height 1 2 h 2 \frac12 h_2 is given by 1 2 2 m g ( h 1 − 1 2 h 2 ) \frac12 2mg\left(h_1-\frac12 h_2\right) .

For the second ball, since it reaches height h 2 h_2 , it takes off with velocity 2 g h 2 \sqrt{2gh_2} . In rising to the heigh 1 2 h 2 \frac12 h_2 , it will lose some kinetic energy to potential energy so that its kinetic energy is 1 2 m 2 g h 2 − 1 2 m g h 2 \frac12 m2gh_2 - \frac12 mgh_2 .

Equating the two kinetic energies gives the simple relationship

h 1 − 1 2 h 2 = 2 h 2 − h 2 h_1 - \frac12 h_2 = 2h_2 - h_2

which leads to h 1 = 3 2 h 2 h1=\frac32 h_2 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I am indebted to you for posting a great solution:-)

Swapnil Das - 5 years, 8 months ago

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